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Bài 3a)
\(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
mà \(a+b=-c\Rightarrow a^3+b^3+c^3=3abc\)
1)a2(b-c)+b2(c-a)+c2(a-b)
=a2b-a2c+b2c-b2a+c2a-c2b
=(a2b-c2b)+(b2c-b2a)+(c2a-a2c)
=b.(a2-c2)-b2.(a-c)-ac.(a-c)
=b.(a-c)(a+c)-b2(a-c)-ac(a-c)
=(a-c)(ab+bc-b2-ac)
=(a-c)[(ab-ac)+(bc-b2)]
=(a-c)[a.(b-c)-b.(b-c)]
=(a-c)(b-c)(a-b)
1.Theo đầu bài ta có:
\(A=x\left(x+2\right)+y\left(y-2\right)-2xy\)
\(=\left(x^2+2x\right)+\left(y^2-2y\right)-2xy\)
\(=\left(x^2+y^2-2xy\right)+\left(2x-2y\right)\)
\(=\left(x-y\right)^2+2\left(x-y\right)\)
Do x - y = 7 nên:
\(=7^2+2\cdot7\)
\(=49+14\)
\(=63\)
Bài 2. Câu 1:
Đặt A = x2 + y2. Khi đó:
\(A-2xy=x^2+y^2-2xy\)
\(\Rightarrow A-2xy=\left(x-y\right)^2\)
Do xy = 4 ; x - y = 3 nên:
\(\Rightarrow A-2\cdot4=3^2\)
\(\Rightarrow A-8=9\)
\(\Rightarrow A=17\)
\(\left(a-b+c\right)^2+\left(c-b\right)^2+2\left(a-b+c\right)\left(b-c\right)\)
\(=\left(a-b+c\right)\left[a-b+c+2\left(b-c\right)\right]+\left(c-b\right)^2\)
\(=\left(a-b+c\right)\left[a-b+c+2b-2c\right]+\left(c-b\right)^2\)
\(=\left(a-b+c\right)\left[a+b-c\right]+\left(c^2-2bc+b^2\right)\)
\(=-c^2+2bc-b^2+a^2\)\(+\left(c^2-2bc+b^2\right)\)
\(=a^2\)
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
Đặt: x2+5x+4=t
Ta có:
\(t\left(t+2\right)-120=t^2+2t-120=t^2+12t-10t-120=t\left(t+12\right)-10\left(t+12\right)\)
\(=\left(t+12\right)\left(t-10\right)=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)
a) (0,5)2.13,7.4-3,7
=\(\left(\frac{1}{2}\right)^2.\frac{137}{10}.4-\frac{37}{10}\)
=\(\frac{1}{4}.\frac{274}{5}-\frac{37}{10}\)
=\(\frac{137}{10}-\frac{37}{10}=\frac{100}{10}=10\)
b) \(\frac{2}{5}.8.\frac{1}{3}+1.\frac{2}{3}.\frac{2}{5}\)
=\(\frac{16}{15}+\frac{4}{15}=\frac{20}{15}=\frac{4}{3}\)
c) \(\left(\frac{3}{4}\right)^2+2.\frac{1}{5}.\left(5,25-\frac{1}{4}\right)-11\)
=\(\frac{9}{16}+\frac{2}{5}\left(\frac{525}{100}-\frac{1}{4}\right)-11\)
=\(\frac{9}{16}+\frac{2}{5}.5-11\)
=\(\frac{9}{16}+2-11=-\frac{135}{16}\)